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Let $(N, h)$ and $(M, g)$ riemannian manifolds with $dim M = dim N$.

We say that M is locally isometric to N if there is a smoother application $F: M\rightarrow N$ such that $$g(v,w)=h(dF_p(v), dF_p(w)),\ \forall v,w \in T_pM,\ \forall p \in M .$$

Another definition is: M is locally isometric to N if for each $p \in M $ exists in an open $U\subset M$ (with $ p \in U $), an open $S\subset N$ and a diffeomorphism $ f : U \rightarrow S $ such that $$g(v,w)=h(df_p(v), df_p(w)),\ \forall v,w \in T_pU,\ \forall p \in U. $$

Of course the first definition implies the second. But these definitions are equivalent? If it is true, how to prove the other implication?

Thanks

Pryscilla Silva
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1 Answers1

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Here is an example showing that both things are not equivalent. Take $M = S^1$ the unit circle with its Riemannian metric (induced by $\mathbb{R}^2$) and let $N=\mathbb{R}$ with its stantard metric. Then $\dim (M) = \dim (N) = 1$ and $M,N$ are locally isometric because of the arclength parameter. But there are no map $F : M \to N$ satisfying $$g(v,w) = h(dF_p(v),dF_p(w))$$ for all $p, v,w \in T_pM$. Indeed, such a map $F$ should be a submersion, but there are no submersion from $S^1$ to $\mathbb{R}$ due to the existence of a critical point e.g. a maximum of $F$.

Holonomia
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